A spring has natural length 20 cm.

Question

If a 25-N force is required to keep it stretched to a length of 30 cm (determining k, the spring constant), how much work is required to stretch it from 20 cm to 25 cm?

Answer 1

(This is really physics) Spring force equals -kx, with x the displacement from natural length. Ignoring the sign and converting cm to meters, if 25 N = k x 0.10 m displacement, then k = 250. Now the work done equates to the change in elastic potential energy with the formula (1/2) k x². Now x = 0.05 m and I get 0.31 Joules.